zoukankan      html  css  js  c++  java
  • ACM学习历程—HDU 5012 Dice(ACM西安网赛)(bfs)

    Problem Description

    There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a1.a2,a3,a4,a5,a6 to be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b1.b2,b3,b4,b5,b6 to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller than 1 and no more than 6 while ai ≠ aj and bi ≠ bj for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

       At the beginning, the two dices may face different(which means there exist some i, ai ≠ bi). Ddy wants to make the two dices look the same from all directions(which means for all i, ai = bi) only by the following four rotation operations.(Please read the picture for more information)


       Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.

    Input

    There are multiple test cases. Please process till EOF.

       For each case, the first line consists of six integers a1,a2,a3,a4,a5,a6, representing the numbers on dice A.

       The second line consists of six integers b1,b2,b3,b4,b5,b6, representing the numbers on dice B.

    Output

     For each test case, print a line with a number representing the answer. If there’s no way to make two dices exactly the same, output -1.

    Sample Input

    1 2 3 4 5 6
    1 2 3 4 5 6
    1 2 3 4 5 6
    1 2 5 6 4 3
    1 2 3 4 5 6
    1 4 2 5 3 6

    Sample Output

    0
    3
    -1

     

    这个题目可以用bfs遍历向前、向后、向左、向右转 ,这样如果用一个数组a[6]记录一种状态,那么最多也只有6!种状态,数量不是很多,可以直接暴力bfs。不过需要记录每个状态是否被访问过。



    代码:

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <cstdlib>
      4 #include <cstring>
      5 #include <cmath>
      6 #include <algorithm>
      7 #include <set>
      8 #include <map>
      9 #include <queue>
     10 #include <string>
     11 #include <vector>
     12 #define inf 0x3fffffff
     13 #define esp 1e-10
     14 using namespace std;
     15 struct node1
     16 {
     17     int dice[6];
     18     int val;
     19 };
     20 struct node
     21 {
     22     node1 qt;
     23     int step;
     24 };
     25 node a;
     26 node1 b;
     27 int bfs()
     28 {
     29     set < int > s;
     30     s.insert(a.qt.val);
     31     queue < node > q;
     32     q.push(a);
     33     while (!q.empty())
     34     {
     35         node f, k;
     36         f = q.front();
     37         q.pop();
     38         if (f.qt.val == b.val) return f.step;
     39         //first
     40         k = f;
     41         swap (k.qt.dice[0], k.qt.dice[5]);
     42         swap (k.qt.dice[4], k.qt.dice[1]);
     43         swap (k.qt.dice[5], k.qt.dice[4]);
     44         k.qt.val = k.qt.dice[0];
     45         for (int y = 1; y < 6; ++y)
     46         {
     47             k.qt.val = 10*k.qt.val + k.qt.dice[y];
     48         }
     49         if (s.find(k.qt.val) == s.end())
     50         {
     51             k.step ++;
     52             q.push(k);
     53             s.insert(k.qt.val);
     54             k.step --;
     55         }
     56         //second
     57         k = f;
     58         swap (k.qt.dice[0], k.qt.dice[5]);
     59         swap (k.qt.dice[4], k.qt.dice[1]);
     60         swap (k.qt.dice[0], k.qt.dice[1]);
     61         k.qt.val = k.qt.dice[0];
     62         for (int y = 1; y < 6; ++y)
     63         {
     64             k.qt.val = 10*k.qt.val + k.qt.dice[y];
     65         }
     66         if (s.find(k.qt.val) == s.end())
     67         {
     68             k.step ++;
     69             q.push(k);
     70             s.insert(k.qt.val);
     71             k.step --;
     72         }
     73         //third
     74         k = f;
     75         swap (k.qt.dice[0], k.qt.dice[2]);
     76         swap (k.qt.dice[1], k.qt.dice[3]);
     77         swap (k.qt.dice[1], k.qt.dice[0]);
     78         k.qt.val = k.qt.dice[0];
     79         for (int y = 1; y < 6; ++y)
     80         {
     81             k.qt.val = 10*k.qt.val + k.qt.dice[y];
     82         }
     83         if (s.find(k.qt.val) == s.end())
     84         {
     85             k.step ++;
     86             q.push(k);
     87             s.insert(k.qt.val);
     88             k.step --;
     89         }
     90         //forth
     91         k = f;
     92         swap (k.qt.dice[0], k.qt.dice[2]);
     93         swap (k.qt.dice[1], k.qt.dice[3]);
     94         swap (k.qt.dice[2], k.qt.dice[3]);
     95         k.qt.val = k.qt.dice[0];
     96         for (int y = 1; y < 6; ++y)
     97         {
     98             k.qt.val = 10*k.qt.val + k.qt.dice[y];
     99         }
    100         if (s.find(k.qt.val) == s.end())
    101         {
    102             k.step ++;
    103             q.push(k);
    104             s.insert(k.qt.val);
    105             k.step --;
    106         }
    107     }
    108     return -1;
    109 }
    110 int main()
    111 {
    112     //freopen ("test.txt", "r", stdin);
    113     while (scanf ("%d", &a.qt.dice[0]) != EOF)
    114     {
    115         for (int i = 1; i < 6; ++i)
    116             scanf ("%d", &a.qt.dice[i]);
    117         a.step = 0;
    118         a.qt.val = a.qt.dice[0];
    119         for (int y = 1; y < 6; ++y)
    120         {
    121             a.qt.val = 10*a.qt.val + a.qt.dice[y];
    122         }
    123         for (int i = 0; i < 6; ++i)
    124             scanf ("%d", &b.dice[i]);
    125         b.val = b.dice[0];
    126         for (int y = 1; y < 6; ++y)
    127         {
    128             b.val = 10*b.val + b.dice[y];
    129         }
    130         printf ("%d
    ", bfs());
    131     }
    132     return 0;
    133 }
    View Code
    把每一道题当作难题去做。
  • 相关阅读:
    php中读取文件内容的几种方法。(file_get_contents:将文件内容读入一个字符串)
    php标准库spl栈SplStack如何使用?
    js进阶正则表达式15验证身份证号(|符号的使用:var reg=/^d{17}[d|X]$|^d{15}$/)(str的方法substr)
    一步一步教你自定义博客园(cnblog)界面
    php中foreach源码分析(编译原理)
    php面试题12(多态web服务器共享session的方法:将session存到数据库)($val=&$data[$key];)
    前端开发必备调试工具(Chrome的F12自带的功能和firebug插件差不多)
    PHP 根据对象属性进行对象数组的排序(usort($your_data, "cmp");)(inside the class: usort($your_data, array($this, "cmp")))
    如何查看一个网页特定效果的js代码(动画效果可js和css)(页面可以看到js的源代码)
    bsh for android : 北京
  • 原文地址:https://www.cnblogs.com/andyqsmart/p/4014507.html
Copyright © 2011-2022 走看看